Question

The point of contact of vertical tangent to the curve given by the equation $ x = 3 - 2\cos \theta $ and $ y = 2 + 3\sin \theta $ is.
 $
  A.\,\,\left( {1,5} \right) \\
  B.\,\,\left( {1,2} \right) \\
  C.\,\,\left( {5,2} \right) \\
  D.\,\,\left( {2,5} \right) \\
  $
Multiple options can be correct.

Answer 1

Hint: Here, two parametric equations. We first differentiate both with respect to $ \theta $ to find their slopes and then equating slope $ \dfrac{{dx}}{{d\theta }} $ equal to zero, as we know that lines which are perpendicular to x-axis or called vertical tangent makes an angle of $ {90^0} $ with x-axis and so $ \tan {90^0} = \infty \Rightarrow \,\,\dfrac{{dx}}{{d\theta }} = 0 $ .

Complete step-by-step answer:
Given that, $ x = 3 - 2\cos \theta $ and $ y = 2 + 3\sin \theta $ are two parametric equations.
We know that slope of tangent is given as: $ \dfrac{{dy}}{{dx}} $
Which can be written as: $ \dfrac{{\dfrac{{dy}}{{d\theta }}}}{{\dfrac{{dx}}{{d\theta }}}} $
Also, for vertical tangents we will take $ \dfrac{{dx}}{{d\theta }} = 0 $
But, we have $ x = 3 - 2\cos \theta $
Therefore, differentiating it with $ \theta $ . We have
 $
  \dfrac{{dx}}{{d\theta }} = \dfrac{d}{{d\theta }}(3) - 2\dfrac{d}{{d\theta }}(\cos \theta ) \\
   \Rightarrow \dfrac{{dx}}{{d\theta }} = - 2\left( { - \sin \theta } \right) \\
   \Rightarrow \dfrac{{dx}}{{dx}} = 2\sin \theta \\
  $
Then, for vertical tangent we have:
 $
  2\sin \theta = 0 \\
   \Rightarrow \sin \theta = 0 \\
  $
By general solution of sine we have,
 $ \theta = 0\,\,or\,\,n\pi $
Substituting value of $ \theta $ in given equations we have:
 $ x = 3 - 2 \cos \theta $
$ \Rightarrow x = 3 - 2 \cos \left( {n\pi } \right) $
$ \Rightarrow x =\text 5 (or ) 1 \left\{ {\cos n \pi = \text 1 (or) - 1} \right\} $
Also, for
 $ y = 2 + 3\sin \theta $
 $ y = 2 + 3\sin \left( {n\pi } \right) $
$ y = 2 + 3(0) $
$ y = 2 \left\{ {\because \sin n \pi = 0} \right\} $
Therefore, from above we see that required points are $ \left( {5,2} \right)\,\,and\,\left( {1,2} \right) $ .
So, the correct answer is “Option B AND C”.

Note: Tangent is a line which touch the curve only at one point and this point is called point of contact and also for vertical tangents are tangents which are perpendicular to x-axis therefore for these we always take $ \dfrac{{dx}}{{d\theta }} = 0 $ and for horizontal tangents we always take $ \dfrac{{dy}}{{d\theta }} = 0 $ and these are always parallel to x axis.